I am looking for a short proof that ∫0∞(sin⁡xx)2dx=π2 What do you think?

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I am looking for a short proof that  ∫0∞(sin⁡xx)2dx=π2  What do you think?

twineg4 · Accepted Answer

Well, it&#039;s not hard to reduce this integral to f(x)=max{0,1−|x|}. It is easy to calculate the Fourier transformf^(ξ)=∫−∞∞f(x)e−ixξ=(sin⁡(ξ2)ξ2)2Taking the inverse Fourier transform, we get∫−∞∞(sin⁡(ξ2)ξ2)2eixξdξ=2πf(x),and the result follows.The second integral can be computed in a similar way. Just take f(x)=χ−11(x) (the indicator function of the interval [−1,1])

Karen Robbins · Accepted Answer

∫0∞sin2xx2dx=∫0∞121−cos⁡2xx2dx  =∫0∞1−cos⁡xx2dx  =12∫−∞∞1−cos⁡xx2dx  =12∫−∞∞1−cos⁡xx2dx  =12∫−∞∞R1−e{ix}x2dx  =12∫−∞∞R1−eix+ix1+x2x2dx  =12R∫−∞∞1−eix+ix1+x2x2dx  Now close the contour in the upper half plane, enclosing the pole at x=i with residue 12i, yielding  ∫0∞sin2xx2dx=12⋅2π⋅12i=π2

Vasquez · Accepted Answer

From squaring the identity sin⁡nxsin⁡x=einx−e−inxeix−e−ix=∑k=0n−1e(2k−n+1)ix and integrating we get nπ=∫−π/2π/2sin2⁡nxsin2⁡xdx Let In=∫−π/2π/2sin2⁡nxnx2dx=∫−nπ/2nπ/2sin2⁡yy2dy Then π−In=1n∫−π/2π/2sin2⁡nx(csc2⁡x−x−2x)dx and so |π−In|≤1n∫−π/2π/2|csc2⁡x−x−2|dx=O(1/n) as x↦csc2⁡x−x−2 extends to a continuous function on [−π/2,π/2]. Hence In→π as n→∞ and π=∫∞∞sin2⁡yy2dy

I am looking for a short proof that \int_0^{\infty}(\frac{\sin x}{x})^2dx=\frac{\pi}{2} What do

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New Questions in Calculus 1